poj3723Conscription

来源：互联网发布：阿里云青岛服务器地址编辑：程序博客网时间：2024/06/06 12:36

链接：http://poj.org/problem?id=3723

题意：n个女生m个男生，r对男女之间的关系d，招募一个人要花费10000，但是如果军营中已经有认识的人那么只需要10000-d。

分析：最大生成树。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=50010;const int mod=1000000007;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=1000000007;const int INF=1000000010;const ll MAX=1ll<<55;const double eps=1e-5;const double inf=~0u>>1;const double pi=acos(-1.0);typedef double db;typedef long double ldb;typedef unsigned int uint;typedef unsigned long long ull;struct node {    int x,y,z;    bool operator < (const node a) const {        return z>a.z;    }}a[N];int f[N];int find_f(int a) {    return f[a]==a?a:f[a]=find_f(f[a]);}int unio(int a,int b) {    int fa=find_f(a),fb=find_f(b);    if (fa!=fb) { f[fa]=fb;return 1; }    else return 0;}int main(){    int i,t,n,m,r,tot,ans;    scanf("%d",&t);    while (t--) {        scanf("%d%d%d", &n, &m, &r);        for (tot=n+m,i=1;i<=tot;i++) f[i]=i;        for (i=1;i<=r;i++) scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z);        sort(a+1,a+r+1);        ans=tot*10000;        for (i=1;i<=r;i++)        if (unio(a[i].x+1,a[i].y+n+1)) ans-=a[i].z;        printf("%d\n", ans);    }    return 0;}

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